"Ionization"

[MedPR]

What exactly does ionization mean.

Consider reaction XY+H2O--->HXY+ +OH-

I know that if I saturate the reaction with water, more products will be formed due to le chatelier's. I also understand that ions are just charged things, so ionization means adding a charge (usually removal or addition of electrons, right?).

My two questions are:

1. If I said something was ionized, how would you know if electrons were added, or if electrons were removed?

2. If the explanation to that reaction example above was either "The degree of ionization will be greater, releasing more OH-" or "The degree of ionization will be less, releasing more OH-" how would you be able to pick one? If the degree of ionization is more, then that means more products since only the products are ions. But if the degree of ionization is less, that means there are less products and by le chatelier's, more products would be formed.

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[milski]

1. You would not know. For some things you might be able to infer it: ionized Na is pretty much guaranteed to be Na+.

No time for 2, somebody else can discuss.


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[chiddler]

I don't understand your second question. could you rephrase, please?

 

[MedPR]

I don't understand your second question. could you rephrase, please?

Sure.

From the reaction in the OP, you can see that the products are ionized, whereas the reactants are not. Adding water to that reaction results in moving the reaction forward, according to lechatelier's principle. The two descriptions (two answer choices) were "The degree of ionization will be greater, releasing more OH-" and '"The degree of ionization will be less, releasing more OH-". I don't understand how you can pick one over the other. I can justify both answers being correct, but obviously only one is really correct.

So, here's how I justify "The degree of ionization will be less, releasing more OH-"

1: If the only ions present are products, then a lower degree of ionization (in the reaction) means that there are less products and more reactants. All else equal, this will cause a shift in the reaction to produce more products. OH- is a product, so lower degree of ionization leads to more OH-.

And here's how I justify "The degree of ionization will be greater, releasing more OH-"

2. If the degree of ionization is greater, that means the excess water is having a greater ionization effect on reactant XY, which means it is producing a lot of ions, and since ions are products, the excess H2O is producing more products; more OH-.

 

[chiddler]

So your response to "The degree of ionization will be less, releasing more OH-" is that there are less ions in solution and therefore products are favored.

This logic seems faulty because you're forgetting that it says *degree* of ionization. Not quantity of ions. Adding water increases degree of ionization for sure. The second reason that you wrote seems correct though: water is added which facilitates ionization and therefore more product.

 

[phaloz]

Ionization is the process of converting an atom or molecule into an ion by adding or removing charged particles such as electrons or other ions.

You can infer that electrons were added if the charge of the ion is less than that of the original species. For example, when Cl ionizes it generally becomes Cl- (gainselectron). Cl has a charge of +0 and Cl- has a charge of -1, therefore electrons were added. The opposite is true when you remove electrons, such as when Mg ionizes to become Mg2+. Also on a side note, electrons are not the only particles that can ionize something. For example, NH4+ can ionize by losing a proton (H+) to form NH3+.

Ok now for 2:

The forward reaction XY + H2O is the ionization reaction. We know this because according to the definition of ionization, something ionizes when it is turned into ions by removing or adding charged particles.

For this reaction XY + H20 ----> HXY+ + OH-
You can clearly see that the left side contains no ions/ionic (there are no charged particles) and right side does (HYX+ and OH-). If you add more H20 it increases ionization because the equilibrium will shift to the right producing more OH-.

 

[gettheleadout]

So your response to "The degree of ionization will be less, releasing more OH-" is that there are less ions in solution and therefore products are favored.

This logic seems faulty because you're forgetting that it says *degree* of ionization. Not quantity of ions. Adding water increases degree of ionization for sure. The second reason that you wrote seems correct though: water is added which facilitates ionization and therefore more product.

I agree with Chiddler here. "the degree of ionization will be greater, releasing more OH-" sounds correct to me. I don't see any way for the other choice to be true. Degree of ionization is a descriptive measure, not something you need to evaluate through Le Chatelier's principle.

And phaloz, NH4+ definitely doesn't ionize to NH3+, as this would mean losing an H-radical. NH4+ gains two electrons from the broken N-H bond as it releases H+, meaning electrons are involved even here as it becomes a neutral NH3. If you ionize, you have electrons moving.

 

[phaloz]

I agree with Chiddler here. "the degree of ionization will be greater, releasing more OH-" sounds correct to me. I don't see any way for the other choice to be true. Degree of ionization is a descriptive measure, not something you need to evaluate through Le Chatelier's principle.

And phaloz, NH4+ definitely doesn't ionize to NH3+, as this would mean losing an H-radical. NH4+ gains two electrons from the broken N-H bond as it releases H+, meaning electrons are involved even here as it becomes a neutral NH3. If you ionize, you have electrons moving.

I see what you're saying. But if I remember correctly, the H+ can be essentially pulled off by another species without any movement of electrons from one to the other. This phenomenon can exemplified by definition of a bronsted acid - which states that something is a classified as a bronsted acid by it's ability to 'donate' a hydrogen cation (H+).
Let's take for example:
NH4+ + B- ----> HB + NH3
In this straightforward acid-base reaction, Ammonium reacts with the bronsted base (B-) to form HB + NH3. The H+ cation is simply pulled off the ammonium. The ammonium readily gives it up because it wants to become a neutral compound (NH3).

 

[gettheleadout]

I see what you're saying. But if I remember correctly, the H+ can be essentially pulled off by another species without any movement of electrons from one to the other. This phenomenon can exemplified by definition of a bronsted acid - which states that something is a classified as a bronsted acid by it's ability to 'donate' a hydrogen cation (H+).
Let's take for example:
NH4+ + B- ----> HB + NH3
In this straightforward acid-base reaction, Ammonium reacts with the bronsted base (B-) to form HB + NH3. The H+ cation is simply pulled off the ammonium. The ammonium readily gives it up because it wants to become a neutral compound (NH3).

The NH4+ consists of a nitrogen atom covalently bonded to four hydrogen atoms (four covalent N-H bonds) and no lone pairs. NH3 (the conjugate base of NH4+) has three covalent N-H bonds and one lone pair. Where do you think that lone pair came from? The H+ ion removed when NH4+ acts as a Bronsted acid doesn't simply exist in the molecule as a cation and then leave, it exists with a complete duet (fulfilled by the covalent bond) and when that bond is broken BOTH of the electrons go to the nitrogen atom (instead of being shared between both N and H) and the H becomes electron-deficient, gaining its positive charge. Any movement of atoms between molecules represents simultaneous movement of electrons, even if it is within the molecules themselves. There is no way for an atom to just "come off," and definitely no way for anything to ionize (gain a formal charge, which is by definition determined by number of valence *electrons*) without electron movement.

In fact, the distinction here (between intermolecular movement of electrons and only intramolecular movement of electrons) is essentially the distinction between redox and acid-base reaction types. I now see that you were referring to intermolecular movement of particles specifically, but it must be noted that intermolecular movement of electrons is still involved.

 

[phaloz]

Alright, I see what you mean. I was looking at from a purely intermolecular aspect, and you further broke it down to the intramolecular mechanisms that are involved/behind the intermolecular phenomenon. Thanks for the clear and concise explanation!